Mean-Field Critical Exponents from Van der Waals Equation

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Hello! Welcome back to our study of the Van der Waals model.

In the last lesson, we derived the Law of Corresponding States. By expressing the Van der Waals equation in terms of reduced variables ( $p_{r}, v_{r}, t_{r}$ ), we obtained a universal equation of state, free of any substance-specific parameters. This demonstrated the powerful concept of universality: near the critical point, the behavior of fluids becomes independent of their microscopic details.

Today, we will quantify this universal behavior. Our goal is to extract the mean-field critical exponents ( $β, γ, δ$ ) from the Van der Waals equation near the critical point. These exponents are universal numbers that describe the characteristic power-law behavior of physical quantities as a phase transition is approached. This analysis will give us our first quantitative look at the nature of critical phenomena and establish a benchmark—the mean-field prediction—against which we will compare more sophisticated theories and experimental results later in the course.

What are Critical Exponents?

Near a critical point, many thermodynamic quantities exhibit singular behavior, meaning they diverge or go to zero as a power law of the reduced temperature, $t = (T - T_{c}) / T_{c}$ , or some other relevant field. Critical exponents characterize these power laws. For the liquid-gas transition, we are interested in three key exponents:

$β$ : Describes how the order parameter vanishes as we approach the critical point from the ordered (two-phase) region. For the liquid-gas system, the order parameter is the difference in density between the liquid and gas phases, $ρ_{l} - ρ_{g}$ .
$ρ_{l} - ρ_{g} \propto (- t)^{β} for t < 0$
$δ$ : Describes the shape of the critical isotherm ( $T = T_{c}$ ). It relates the pressure difference from the critical pressure to the volume (or density) difference from the critical volume.
$P - P_{c} \propto ∣ V_{m} - V_{m, c} ∣^{δ} for t = 0$
$γ$ : Describes the divergence of the isothermal compressibility, $K_{T}$ , as we approach the critical point from the disordered (single-phase) region.
$K_{T} = - \frac{1}{V _{m}} (\frac{\partial V _{m}}{\partial P})_{T} \propto t^{- γ} for t > 0$

Our task is to find the values of $β$ , $γ$ , and $δ$ predicted by the Van der Waals model.

Expanding Around the Critical Point

Our starting point is the Law of Corresponding States we derived previously:
$p_{r} = \frac{8 t _{r}}{3 v _{r} - 1} - \frac{3}{v _{r}^{2}}$
To analyze the behavior near the critical point, we introduce small deviation variables from the critical point coordinates $(p_{r}, v_{r}, t_{r}) = (1, 1, 1)$ :

$π = p_{r} - 1$
$ω = v_{r} - 1$
$τ = t_{r} - 1$

Our goal is to find an approximate relation $π (ω, τ)$ for small $ω$ and $τ$ . We can achieve this by substituting $p_{r} = 1 + π$ , $v_{r} = 1 + ω$ , and $t_{r} = 1 + τ$ into the reduced equation and performing a Taylor series expansion.

The substitution gives:
$1 + π = \frac{8 ( 1 + τ )}{3 ( 1 + ω ) - 1} - \frac{3}{( 1 + ω ) ^{2}} = \frac{8 ( 1 + τ )}{2 + 3 ω} - \frac{3}{( 1 + ω ) ^{2}}$
Expanding the denominators using the geometric series $1/ (1 + x) \approx 1 - x + x^{2} - ...$ and keeping terms up to third order in the small variables (as we anticipate needing $ω^{3}$ for the exponent $δ$ ), we find:
$π \approx 4 τ - 6 τ ω - \frac{3}{2} ω^{3} + O (τ ω^{2}, ω^{4})$
This expanded equation is the master key for deriving all three exponents.

Test your understanding!

The Taylor expansion of a function $f (x, y)$ around $(0, 0)$ is $f (x, y) \approx f (0, 0) + x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} + \frac{1}{2} x^{2} \frac{\partial ^{2} f}{\partial x ^{2}} + ...$ .
In our case, $π (ω, τ) = p_{r} (1 + ω, 1 + τ) - 1$ . The absence of an $ω^{2}$ term in our expansion is significant. Why is the coefficient of $ω^{2}$ exactly zero?

Show answer

The coefficient of the $ω^{2}$ term in the expansion is proportional to $(\frac{\partial ^{2} p _{r}}{\partial v _{r}^{2}})$ evaluated at the critical point. We defined the critical point as the inflection point on the critical isotherm where both the first and second derivatives of pressure with respect to volume are zero:
$(\frac{\partial p _{r}}{\partial v _{r}})_{v_{r} = 1, t_{r} = 1} = 0 and (\frac{\partial ^{2} p _{r}}{\partial v _{r}^{2}})_{v_{r} = 1, t_{r} = 1} = 0$
The vanishing of the first derivative explains why there's no term linear in $ω$ (with a constant coefficient). The vanishing of the second derivative is precisely why the $ω^{2}$ term is absent. The first non-vanishing pure derivative with respect to volume is the third, which gives rise to the $ω^{3}$ term.

Deriving the Critical Exponents

With the expanded equation of state, we can now extract the exponents one by one. The following document provides a detailed walkthrough of these derivations. We will follow its logic closely.

[PDF] van der Waals gas

The following document, 'van der Waals gas' from Binghamton University, performs the explicit calculations for the critical exponents. We will use it as our guide.

Please skim through section 6, 'Critical points and critical exponents' (starting around page 18). We will work through the derivations for exponents β, γ, and δ step-by-step, and you can refer to subsections (a), (b), and (d) for the specific algebraic details.

1. The Exponent $β$ (Coexistence Curve)

To find $β$ , we look at the system just below the critical temperature ( $τ < 0$ ), where the liquid and gas phases coexist. This means there are two different volumes, $ω_{l}$ (liquid) and $ω_{g}$ (gas), that have the same pressure $π$ .

From our expansion $π \approx 4 τ - 6 τ ω - \frac{3}{2} ω^{3}$ , the Maxwell construction requires that the coexistence pressure $π$ is such that the two stable volume solutions enclose equal areas with the unphysical part of the isotherm. Due to the symmetry of the cubic equation around its inflection point, this condition is met when we set $π$ to the pressure at $ω = 0$ , i.e., $π \approx 4 τ$ .

Substituting this into the expansion gives:
$4 τ = 4 τ - 6 τ ω - \frac{3}{2} ω^{3}$
$- 6 τ ω - \frac{3}{2} ω^{3} = 0 ⟹ ω (- 6 τ - \frac{3}{2} ω^{2}) = 0$
This equation has three solutions for $ω$ :

$ω = 0$ (the unstable state in the middle)
$ω^{2} = - 4 τ$

Since $τ < 0$ , the other two solutions are real: $ω_{g} = 2 - τ$ and $ω_{l} = - 2 - τ$ .
The order parameter is the difference in density. For small $ω$ , the reduced density is $ρ_{r} = 1/ v_{r} = 1/ (1 + ω) \approx 1 - ω$ .
$ρ_{r, l} - ρ_{r, g} \approx (1 - ω_{l}) - (1 - ω_{g}) = ω_{g} - ω_{l} = 4 - τ$
The order parameter scales as $(- τ)^{1/2}$ . Comparing this to the definition $Δ ρ \propto (- τ)^{β}$ , we find:
$β = \frac{1}{2}$

2. The Exponent $δ$ (Critical Isotherm)

To find $δ$ , we analyze the system at the critical temperature, so $τ = 0$ . Our master expansion simplifies dramatically:
$π \approx - \frac{3}{2} ω^{3}$
This equation relates the pressure deviation to the volume deviation along the critical isotherm. By definition, $π \propto ω^{δ}$ . We can immediately read off the exponent:
$δ = 3$
This derivation is particularly elegant. As we reasoned in the "Test your understanding" box, the physics of the critical point (being an inflection point) forces the relationship between pressure and volume to be cubic at leading order.

3. The Exponent $γ$ (Isothermal Compressibility)

To find $γ$ , we look at the isothermal compressibility $K_{T}$ as we approach the critical point from above ( $τ > 0$ ) along the critical isochore ( $ω = 0$ ).
The reduced isothermal compressibility is proportional to $- (\partial v_{r} / \partial p_{r})_{t_{r}} = - (\partial ω / \partial π)_{τ}$ . It is easier to first compute its inverse, $(\partial π / \partial ω)_{τ}$ .

Differentiating our master expansion with respect to $ω$ :
$(\frac{\partial π}{\partial ω})_{τ} \approx - 6 τ - \frac{9}{2} ω^{2}$
Now, we evaluate this on the critical isochore, so we set $ω = 0$ :
$(\frac{\partial π}{\partial ω})_{τ, ω = 0} \approx - 6 τ$
The compressibility is the negative reciprocal:
$K_{T, r} \propto - (\frac{\partial ω}{\partial π})_{τ, ω = 0} \approx - \frac{1}{- 6 τ} = \frac{1}{6 τ}$
The compressibility diverges as $τ^{- 1}$ . Comparing this to the definition $K_{T} \propto τ^{- γ}$ , we find:
$γ = 1$

The Mean-Field Exponents

We have successfully derived the three critical exponents for the Van der Waals model:

$β = 1/2$
$γ = 1$
$δ = 3$

These are known as the classical or mean-field critical exponents. The term "mean-field" is crucial. The Van der Waals equation is fundamentally a mean-field theory because it replaces the complex, fluctuating interactions a particle experiences with an average (or "mean") interaction field from all other particles. This averaging washes out the intricate correlations and fluctuations that become dominant near a critical point.

The remarkable thing, which you will see in future lessons, is that any theory based on a mean-field approximation—such as the mean-field theory of the Ising model or the basic Landau theory of phase transitions—yields these exact same exponents.

Statistical Physics by David Tong

For a preview of this universality, you can read the following section from David Tong's notes. It derives the same exponents for the mean-field Ising model, highlighting that the results are identical to what we just found for the Van der Waals gas.

I recommend reading the section 'Critical Exponents' (Section 5.2.2). The prior section (5.1.4) derives the vdW exponents with a slightly different but elegant line of reasoning. Section 5.2.2 then computes them for magnetism and explicitly states the equivalence. This powerfully reinforces the idea of universality among mean-field theories.

This universality is a cornerstone of the modern understanding of phase transitions. The exponents depend not on the microscopic details of the system (e.g., liquid-gas vs. magnet) but on fundamental properties like the dimensionality of space and the symmetries of the order parameter. The mean-field exponents provide a universal baseline, which is quantitatively accurate for systems above a certain "upper critical dimension" ( $d \geq 4$ ) and serves as the starting point for more advanced theories like the Renormalization Group, which we will explore later.

Conclusion

In this lesson, we quantified the universal behavior of the Van der Waals fluid near its critical point.

Key Takeaways:

By performing a Taylor expansion of the reduced equation of state around the critical point, we derived an effective equation, $π \approx 4 τ - 6 τ ω - \frac{3}{2} ω^{3}$ , that governs the physics in this region.
From this expansion, we extracted the critical exponents that describe the singular behavior of the order parameter ( $β$ ), the critical isotherm ( $δ$ ), and the compressibility ( $γ$ ).
The values we found, $β = 1/2$ , $γ = 1$ , and $δ = 3$ , are the universal mean-field critical exponents. They are a hallmark of any theory that neglects critical fluctuations.

Preview of the Next Lesson:
So far, we have mostly ignored the S-shaped, "unphysical" parts of the Van der Waals isotherms below $T_{c}$ , using the Maxwell construction to bypass them. In our next lesson, we will explore the physical significance of these regions. We will discuss the concepts of metastability and nucleation, which correspond to the states of superheated liquids and supercooled gases predicted by the model. This will give us a more complete picture of the rich non-equilibrium phenomena surrounding a first-order phase transition.